Problem: A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.

Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).

Output

Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Examples

Input

1 0


Output

1 2

Input

2 1


Output

3 2 1 4


Input

4 0


Output

2 7 4 6 1 3 5 8

Note

Record |x| represents the absolute value of number x. 

In the first sample |1 - 2| - |1 - 2| = 0.

In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.

In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int toInt(string s) {
  istringstream sin(s);
  int t;
  sin >> t;
  return t;
}
template <class T>
string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
long long toInt64(string s) {
  istringstream sin(s);
  long long t;
  sin >> t;
  return t;
}
template <class T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
long long C(long long n, long long m) {
  long long a[100], b[100];
  if (n < m) return 0;
  if (m == 0) return 1;
  if (m == 1) return n;
  for (int i = 1; i <= m; i++) {
    a[i] = n - i + 1;
    b[i] = i;
  }
  long long ans = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) {
      long long s = gcd(a[i], b[j]);
      if (s > 1) {
        a[i] /= s;
        b[j] /= s;
      }
    }
    ans *= a[i];
  }
  return ans;
}
const int N = 105;
struct node {
  int x, p;
} p[N];
bool cmp(const node &a, const node &b) { return a.x < b.x; }
int d[1000005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    d[i * 2 - 1] = i * 2;
    d[i * 2] = i * 2 - 1;
  }
  for (int i = 1; i <= k; i++) {
    int t = d[i * 2 - 1];
    d[i * 2 - 1] = d[i * 2];
    d[i * 2] = t;
  }
  for (int i = 1; i <= n * 2; i++)
    if (i > 1)
      cout << " " << d[i];
    else
      cout << d[i];
  cout << endl;
  return 0;
}